设数列{an}满足a1+2a2=3,且对任意的n∈N*,点Pn(n,an)都有PnPn+1=(1,2),则{an}的前n项和Sn为()A.n(n-43)B.n(n-34)C.n(n-23)D.n(n-

题目简介

设数列{an}满足a1+2a2=3,且对任意的n∈N*,点Pn(n,an)都有PnPn+1=(1,2),则{an}的前n项和Sn为()A.n(n-43)B.n(n-34)C.n(n-23)D.n(n-

题目详情

设数列{an}满足a1+2a2=3,且对任意的n∈N*,点Pn(n,an)都有
PnPn+1
=(1,2)
,则{an}的前n项和Sn为(  )
A.n(n-
4
3
)
B.n(n-
3
4
)
C.n(n-
2
3
)
D.n(n-
1
2
)
题型:单选题难度:中档来源:东城区二模

答案

∵Pn(n,an),∴Pn+1(n+1,an+1),故
PnPn+1
=(1,an+1-an)  =(1,2)

an+1-an=2,∴an是等差数列,公差d=2,将a2=a1+2,代入a1+2a2=3中,
解得a1=-class="stub"1
3
,∴an=-class="stub"1
3
+2(n-1)=2n-class="stub"7
3

Sn=
a1+an
2
n=
-class="stub"1
3
+2n-class="stub"7
3
2
n=(n-class="stub"4
3
)n

故选A.

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